Unit 5, lesson 63

Sum and Difference Formulas

How to find the sine, cosine and tangent of a sum or difference of two angles from the values of each angle, why the formulas hold, and how they give exact values such as sin 75°.

  • Theory part 125
  • Practice part 126
  • Identities and Formulas

No special triangle has a 75∘75^\circ angle, yet 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ, and both of those angles have exact sines and cosines. The sum and difference formulas say how to combine them. They also settle a common mistake, because the sine of a sum is not the sum of the sines, and calculus uses the sine sum formula to find the derivative of sine.

Key terms

Difference formula for cosine

Definition For all angles AA and BB, both measured in radians or both in degrees,

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B.\begin{aligned} \cos(A - B) &= \cos A \cos B \\[0.4em] &\quad + \sin A \sin B. \end{aligned}

The reason is a distance on the unit circle. The points P=(cos⁡A,sin⁡A)P = (\cos A, \sin A) and Q=(cos⁡B,sin⁡B)Q = (\cos B, \sin B) lie on the unit circle, and the angle from QQ to PP at the origin is A−BA - B. Rotating both points about the origin by −B-B moves QQ to (1,0)(1, 0) and PP to (cos⁡(A−B),sin⁡(A−B))\big(\cos(A - B), \sin(A - B)\big) without changing their distance. Expand the squares and use sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1 and sin⁡2B+cos⁡2B=1\sin^2 B + \cos^2 B = 1. The squared distance before the rotation is

(cos⁡A−cos⁡B)2+(sin⁡A−sin⁡B)2=2−2(cos⁡Acos⁡B+sin⁡Asin⁡B),\begin{gathered} (\cos A - \cos B)^2 \\[0.4em] {} + (\sin A - \sin B)^2 \\[0.4em] = 2 - 2(\cos A \cos B + \sin A \sin B), \end{gathered}

and after it is

(cos⁡(A−B)−1)2+sin⁡2(A−B)=2−2cos⁡(A−B).\begin{gathered} \big(\cos(A - B) - 1\big)^2 + \sin^2(A - B) \\[0.4em] = 2 - 2\cos(A - B). \end{gathered}

Setting the two equal gives the formula.

In plain words The cosine of a difference is not the difference of the cosines. Multiply cosine by cosine and sine by sine, then add.

Example With A=45∘A = 45^\circ and B=30∘B = 30^\circ,

cos⁡15∘=cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘=22⋅32+22⋅12=6+24.\begin{aligned} \cos 15^\circ &= \cos 45^\circ \cos 30^\circ \\[0.4em] &\quad + \sin 45^\circ \sin 30^\circ \\[0.4em] &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\[0.8em] &= \frac{\sqrt{6} + \sqrt{2}}{4}. \end{aligned}

A calculator gives cos⁡15∘≈0.9659\cos 15^\circ \approx 0.9659, and 6+24≈0.9659\frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659 as well.

Sum formula for cosine

Definition For all angles AA and BB,

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B.\begin{aligned} \cos(A + B) &= \cos A \cos B \\[0.4em] &\quad - \sin A \sin B. \end{aligned}

It follows from the difference formula with A+B=A−(−B)A + B = A - (-B), using cos⁡(−B)=cos⁡B\cos(-B) = \cos B and sin⁡(−B)=−sin⁡B\sin(-B) = -\sin B.

In plain words Same products as before, with the sign between them flipped. A plus sign in the angle gives a minus sign in the formula.

Example The angle 7π12\frac{7\pi}{12} equals π3+π4\frac{\pi}{3} + \frac{\pi}{4}, so

cos⁡7π12=cos⁡π3cos⁡π4−sin⁡π3sin⁡π4=12⋅22−32⋅22=2−64≈−0.2588.\begin{aligned} \cos \frac{7\pi}{12} &= \cos \frac{\pi}{3} \cos \frac{\pi}{4} - \sin \frac{\pi}{3} \sin \frac{\pi}{4} \\[0.8em] &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\[0.8em] &= \frac{\sqrt{2} - \sqrt{6}}{4} \approx -0.2588. \end{aligned}

The answer is negative, as it must be, since 7π12\frac{7\pi}{12} is 105∘105^\circ, in the second quadrant.

Cofunction identities

Definition For every real θ\theta, in radians,

cos⁡(π2−θ)=sin⁡θ,sin⁡(π2−θ)=cos⁡θ.\begin{gathered} \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta, \\[0.8em] \sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta. \end{gathered}

In degrees, replace π2\frac{\pi}{2} by 90∘90^\circ. The first identity is the cosine difference formula with A=π2A = \frac{\pi}{2} and B=θB = \theta, since

cos⁡(π2−θ)=0⋅cos⁡θ+1⋅sin⁡θ=sin⁡θ.\begin{aligned} \cos\left(\frac{\pi}{2} - \theta\right) &= 0 \cdot \cos \theta + 1 \cdot \sin \theta \\[0.8em] &= \sin \theta. \end{aligned}

The second is the first applied to the angle π2−θ\frac{\pi}{2} - \theta in place of θ\theta. Dividing the second by the first gives, wherever sin⁡θ≠0\sin \theta \neq 0,

tan⁡(π2−θ)=cot⁡θ.\tan\left(\frac{\pi}{2} - \theta\right) = \cot \theta.

In plain words The sine of an angle is the cosine of its complement. In a right triangle the side opposite one acute angle is the side adjacent to the other one.

Example The acute angles of a right triangle add up to 90∘90^\circ. If one of them is 35∘35^\circ, the other is 55∘55^\circ, and sin⁡35∘=cos⁡55∘≈0.5736\sin 35^\circ = \cos 55^\circ \approx 0.5736.

Sum formula for sine and its cofunction proof

Definition For all angles AA and BB,

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B.\begin{aligned} \sin(A + B) &= \sin A \cos B \\[0.4em] &\quad + \cos A \sin B. \end{aligned}

The cofunction proof turns the sine into a cosine and uses the cosine difference formula, with the angles π2−A\frac{\pi}{2} - A and BB.

sin⁡(A+B)=cos⁡(π2−(A+B))=cos⁡((π2−A)−B)=cos⁡(π2−A)cos⁡B+sin⁡(π2−A)sin⁡B=sin⁡Acos⁡B+cos⁡Asin⁡B.\begin{aligned} \sin(A + B) &= \cos\left(\frac{\pi}{2} - (A + B)\right) \\[0.8em] &= \cos\left(\left(\frac{\pi}{2} - A\right) - B\right) \\[0.8em] &= \cos\left(\frac{\pi}{2} - A\right) \cos B \\[0.8em] &\quad + \sin\left(\frac{\pi}{2} - A\right) \sin B \\[0.8em] &= \sin A \cos B \\[0.4em] &\quad + \cos A \sin B. \end{aligned}

Together with the cosine sum formula this has a geometric meaning. Rotating the point (cos⁡A,sin⁡A)(\cos A, \sin A) counterclockwise by the angle BB lands on (cos⁡(A+B),sin⁡(A+B))\big(\cos(A + B), \sin(A + B)\big), and the two formulas give the two coordinates of the rotated point.

In plain words The sine of a sum mixes the two functions, sine times cosine plus cosine times sine. The cofunction identities let the cosine formula do all the work.

Example With A=45∘A = 45^\circ and B=30∘B = 30^\circ,

sin⁡75∘=22⋅32+22⋅12=6+24,\begin{aligned} \sin 75^\circ &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\[0.8em] &= \frac{\sqrt{6} + \sqrt{2}}{4}, \end{aligned}

the same number as cos⁡15∘\cos 15^\circ, as the cofunction identity predicts. Notice that adding the sines would be wrong. For instance sin⁡(30∘+60∘)=1\sin(30^\circ + 60^\circ) = 1, while sin⁡30∘+sin⁡60∘≈1.366\sin 30^\circ + \sin 60^\circ \approx 1.366.

Difference formula for sine

Definition For all angles AA and BB,

sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B.\begin{aligned} \sin(A - B) &= \sin A \cos B \\[0.4em] &\quad - \cos A \sin B. \end{aligned}

It comes from the sum formula with −B-B in place of BB. Unlike the cosine formulas, the order matters here, because swapping AA and BB changes the sign, sin⁡(B−A)=−sin⁡(A−B)\sin(B - A) = -\sin(A - B).

In plain words Same pattern as the sine sum formula with a minus sign between the products. Keep sin⁡Acos⁡B\sin A \cos B first.

Example The angle π12\frac{\pi}{12} equals π4−π6\frac{\pi}{4} - \frac{\pi}{6}, so

sin⁡π12=22⋅32−22⋅12=6−24≈0.2588.\begin{aligned} \sin \frac{\pi}{12} &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\[0.8em] &= \frac{\sqrt{6} - \sqrt{2}}{4} \approx 0.2588. \end{aligned}

Sum and difference formulas for tangent

Definition If tan⁡A\tan A, tan⁡B\tan B and tan⁡(A+B)\tan(A + B) are all defined, then

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B.\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}.

If tan⁡A\tan A, tan⁡B\tan B and tan⁡(A−B)\tan(A - B) are all defined, then

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B.\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.

Each comes from dividing the sine formula by the cosine formula and then dividing the top and the bottom by cos⁡Acos⁡B\cos A \cos B. When tan⁡A\tan A and tan⁡B\tan B are defined but tan⁡Atan⁡B=1\tan A \tan B = 1, the cosine of A+BA + B is zero and tan⁡(A+B)\tan(A + B) does not exist.

In plain words The tangent of a sum needs only the two tangents. The sign in the numerator matches the sign in the angle, and the sign in the denominator is the opposite one.

Example With tan⁡45∘=1\tan 45^\circ = 1 and tan⁡30∘=33\tan 30^\circ = \frac{\sqrt{3}}{3},

tan⁡75∘=1+331−33=3+33−3=(3+3)29−3=2+3≈3.732.\begin{aligned} \tan 75^\circ &= \frac{1 + \frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}} \\[0.8em] &= \frac{(3 + \sqrt{3})^2}{9 - 3} \\[0.8em] &= 2 + \sqrt{3} \approx 3.732. \end{aligned}

Combining angles from given values

Definition To use a sum or difference formula you need the sine and cosine of both angles. If only one of sin⁡θ\sin \theta and cos⁡θ\cos \theta is given, together with the quadrant of θ\theta, find the other with the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 and choose its sign from the quadrant.

In plain words Fill in the missing sine or cosine first, with the right sign, and only then use the formula.

Example Suppose sin⁡α=35\sin \alpha = \frac{3}{5} with α\alpha in Quadrant II and cos⁡β=513\cos \beta = \frac{5}{13} with β\beta in Quadrant IV. Cosine is negative in Quadrant II and sine is negative in Quadrant IV, so cos⁡α=−45\cos \alpha = -\frac{4}{5} and sin⁡β=−1213\sin \beta = -\frac{12}{13}. Then

sin⁡(α+β)=35⋅513+(−45)(−1213)=15+4865=6365,\begin{aligned} \sin(\alpha + \beta) &= \frac{3}{5} \cdot \frac{5}{13} \\[0.8em] &\quad + \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) \\[0.8em] &= \frac{15 + 48}{65} = \frac{63}{65}, \end{aligned}cos⁡(α+β)=(−45)⋅513−35(−1213)=−20+3665=1665.\begin{aligned} \cos(\alpha + \beta) &= \left(-\frac{4}{5}\right) \cdot \frac{5}{13} \\[0.8em] &\quad - \frac{3}{5}\left(-\frac{12}{13}\right) \\[0.8em] &= \frac{-20 + 36}{65} = \frac{16}{65}. \end{aligned}

As a check, 632+162=4225=65263^2 + 16^2 = 4225 = 65^2, so the two values satisfy the Pythagorean identity.

Check yourself

  1. Write sin⁡50∘cos⁡10∘+cos⁡50∘sin⁡10∘\sin 50^\circ \cos 10^\circ + \cos 50^\circ \sin 10^\circ as a single sine and find its exact value.
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This is the sine sum formula read from right to left, so the expression equals

sin⁡(50∘+10∘)=sin⁡60∘=32.\sin(50^\circ + 10^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}.
  1. Find the exact value of tan⁡π12\tan \frac{\pi}{12}.
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Write π12=π4−π6\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6} and use the tangent difference formula.

tan⁡π12=1−331+33=3−33+3=(3−3)26=2−3.\begin{aligned} \tan \frac{\pi}{12} &= \frac{1 - \frac{\sqrt{3}}{3}}{1 + \frac{\sqrt{3}}{3}} = \frac{3 - \sqrt{3}}{3 + \sqrt{3}} \\[0.8em] &= \frac{(3 - \sqrt{3})^2}{6} = 2 - \sqrt{3}. \end{aligned}
  1. Show that sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x for every real xx.
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The sine difference formula with sin⁡π=0\sin \pi = 0 and cos⁡π=−1\cos \pi = -1 gives

sin⁡(π−x)=sin⁡πcos⁡x−cos⁡πsin⁡x=0+sin⁡x=sin⁡x.\begin{aligned} \sin(\pi - x) &= \sin \pi \cos x \\[0.4em] &\quad - \cos \pi \sin x \\[0.4em] &= 0 + \sin x = \sin x. \end{aligned}

How this lesson connects

Where this leads in Calculus